1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Deriviative of a log

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    find dy/dx of log3*arccsc2^x

    3. The attempt at a solution

    I got up to 1/ln3csc2^x * 1/2^x(sqrt(2^2x-1)

    However there's also a 2^x*ln2 after that

    How is that?

    In otherwords how would I derive 2^x would i treat it like e^x or something?

    Also if its x^x why would that be 1+lnx?
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 23, 2009 #2
    I don't feel like reading your main question; it just looks messy. However, I can answer your sub q I think!

    d/dx of e^x is e^x * ln (e) = e^x and d/dx of 2^x is 2^x ln (2).

    To derive this, suppose y = 2^x, then ln y = ln (2^x) = x ln (2).

    derivative of both sides gives you:
    1/y * dy/dx = ln (2) multiply both sides by y to get:

    dy/dx = y ln (2) and substitute y = 2^x to get

    dy/dx = 2^x ln (2)

    I also did x ^ x the same way, try it before looking at my solution..

    let y = x^x then
    ln y = x ln (x)

    1/y * dy/dx = 1 * ln (x) + 1/x * x = ln (x) + 1 ->

    dy/dx = y ln(x) + y = x^x * ln (x) + x^x or more commonly x^x * [ln (x) + 1]
  4. Feb 23, 2009 #3
    Thanks!! that's all i needed
  5. Feb 23, 2009 #4


    User Avatar
    Science Advisor

    That is log3(arcsec(2x))?

    I have no idea how you got that! How did "arcsec" become "csc"? The derivative of log x is 1/x so you should have 1/(arcsec(2x)) times the derivative of that denominator.

    [tex]2^x= e^{ln(2^x)}= e^{x ln(2)}[/tex] so the derivative of 2x is ln(2)2x.

    To differentiate y= xx, write it as ln(y)= x ln(x). On the left, the derivative is (1/y) y'. On the right, it is ln(x)+ x(1/x)= ln(x)+ 1. So y'= (ln(x)+ 1)y= (ln(x)+ 1)xx.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook